3.1.5 \(\int x (a+b \text {ArcTan}(c x)) \, dx\) [5]

Optimal. Leaf size=37 \[ -\frac {b x}{2 c}+\frac {b \text {ArcTan}(c x)}{2 c^2}+\frac {1}{2} x^2 (a+b \text {ArcTan}(c x)) \]

[Out]

-1/2*b*x/c+1/2*b*arctan(c*x)/c^2+1/2*x^2*(a+b*arctan(c*x))

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Rubi [A]
time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {4946, 327, 209} \begin {gather*} \frac {1}{2} x^2 (a+b \text {ArcTan}(c x))+\frac {b \text {ArcTan}(c x)}{2 c^2}-\frac {b x}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTan[c*x]),x]

[Out]

-1/2*(b*x)/c + (b*ArcTan[c*x])/(2*c^2) + (x^2*(a + b*ArcTan[c*x]))/2

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rubi steps

\begin {align*} \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx &=\frac {1}{2} x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{2} (b c) \int \frac {x^2}{1+c^2 x^2} \, dx\\ &=-\frac {b x}{2 c}+\frac {1}{2} x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {b \int \frac {1}{1+c^2 x^2} \, dx}{2 c}\\ &=-\frac {b x}{2 c}+\frac {b \tan ^{-1}(c x)}{2 c^2}+\frac {1}{2} x^2 \left (a+b \tan ^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 42, normalized size = 1.14 \begin {gather*} -\frac {b x}{2 c}+\frac {a x^2}{2}+\frac {b \text {ArcTan}(c x)}{2 c^2}+\frac {1}{2} b x^2 \text {ArcTan}(c x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcTan[c*x]),x]

[Out]

-1/2*(b*x)/c + (a*x^2)/2 + (b*ArcTan[c*x])/(2*c^2) + (b*x^2*ArcTan[c*x])/2

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Maple [A]
time = 0.11, size = 40, normalized size = 1.08

method	result	size
derivativedivides	\(\frac {\frac {c^{2} x^{2} a}{2}+\frac {\arctan \left (c x \right ) b \,c^{2} x^{2}}{2}-\frac {x b c}{2}+\frac {b \arctan \left (c x \right )}{2}}{c^{2}}\)	\(40\)
default	\(\frac {\frac {c^{2} x^{2} a}{2}+\frac {\arctan \left (c x \right ) b \,c^{2} x^{2}}{2}-\frac {x b c}{2}+\frac {b \arctan \left (c x \right )}{2}}{c^{2}}\)	\(40\)
risch	\(-\frac {i x^{2} b \ln \left (i c x +1\right )}{4}+\frac {i b \,x^{2} \ln \left (-i c x +1\right )}{4}+\frac {a \,x^{2}}{2}-\frac {b x}{2 c}+\frac {b \arctan \left (c x \right )}{2 c^{2}}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctan(c*x)),x,method=_RETURNVERBOSE)

[Out]

1/c^2*(1/2*c^2*x^2*a+1/2*arctan(c*x)*b*c^2*x^2-1/2*x*b*c+1/2*b*arctan(c*x))

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Maxima [A]
time = 0.46, size = 37, normalized size = 1.00 \begin {gather*} \frac {1}{2} \, a x^{2} + \frac {1}{2} \, {\left (x^{2} \arctan \left (c x\right ) - c {\left (\frac {x}{c^{2}} - \frac {\arctan \left (c x\right )}{c^{3}}\right )}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctan(c*x)),x, algorithm="maxima")

[Out]

1/2*a*x^2 + 1/2*(x^2*arctan(c*x) - c*(x/c^2 - arctan(c*x)/c^3))*b

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Fricas [A]
time = 3.43, size = 34, normalized size = 0.92 \begin {gather*} \frac {a c^{2} x^{2} - b c x + {\left (b c^{2} x^{2} + b\right )} \arctan \left (c x\right )}{2 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctan(c*x)),x, algorithm="fricas")

[Out]

1/2*(a*c^2*x^2 - b*c*x + (b*c^2*x^2 + b)*arctan(c*x))/c^2

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Sympy [A]
time = 0.14, size = 42, normalized size = 1.14 \begin {gather*} \begin {cases} \frac {a x^{2}}{2} + \frac {b x^{2} \operatorname {atan}{\left (c x \right )}}{2} - \frac {b x}{2 c} + \frac {b \operatorname {atan}{\left (c x \right )}}{2 c^{2}} & \text {for}\: c \neq 0 \\\frac {a x^{2}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*atan(c*x)),x)

[Out]

Piecewise((a*x**2/2 + b*x**2*atan(c*x)/2 - b*x/(2*c) + b*atan(c*x)/(2*c**2), Ne(c, 0)), (a*x**2/2, True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctan(c*x)),x, algorithm="giac")

[Out]

sage0*x

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Mupad [B]
time = 0.12, size = 34, normalized size = 0.92 \begin {gather*} \frac {a\,x^2}{2}+\frac {b\,\mathrm {atan}\left (c\,x\right )}{2\,c^2}+\frac {b\,x^2\,\mathrm {atan}\left (c\,x\right )}{2}-\frac {b\,x}{2\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atan(c*x)),x)

[Out]

(a*x^2)/2 + (b*atan(c*x))/(2*c^2) + (b*x^2*atan(c*x))/2 - (b*x)/(2*c)

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